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Let $\mathcal{H}$ be the Hilbert space of a quantum system and $A$ one observable in $\mathcal{H}$. If $A$ has discrete spectrum $\{a_n : n \in \mathbb{N}\}$ for simplicity, then by the postulates of Quantum Mechanics, the probability that a measurement of $A$ on the state $|\psi\rangle$ will result $a_n$ is given by

$$P(a_n) = \sum_{i}|\langle \varphi_n^i |\psi\rangle|^2$$

where $|\varphi_n^i\rangle$ are the states with same eigenvalue and $i=1,\dots,g_n$ is the number of degeneracies.

This is all right, but if $U\subset \mathbb{R}$ is a set and we want to find the probability that the measurement will be in the set I don't know, from the postulates, how to proceed.

For example, if the observable is the Hamiltonian $H$ we might want to consider the probbility that the energy is greater or equal to $E_0$. In that case we would like to know if a measurement of energy will give a number in the set $\{\lambda \in \mathbb{R} : \lambda \geq E_0\}$.

In that case, how do we compute this? One intuitive way would be to sum up all the probabilities of the possible measurement which lies in $U$, that is, the sum

$$P(U) = \sum_{n\in J}P(a_n),$$

where $n\in J$ if and only if $a_n \in U$. But I'm unsure this is the way, I can't see how this follows from the postulates.

So if we want the probability that the result of the measurement of one observable lies in a given set of real numbers, how should we do and how this follows from the postulates?

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  • $\begingroup$ It follows, simply, by the definition of probability of union events as sum of the single probabilities (minus the possible pairwise intersections) and it has nothing to do with quantum mechanics, which, in turn, only postulates the single probabilities to be, as you said, the scalar product modulus squares. From then on, it's just the sum. $\endgroup$
    – gented
    Nov 8, 2015 at 0:09

2 Answers 2

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Probabilities of independent outcomes of the same measurement are always additive, which means that the expression $$P(U) = \sum_{n\in J}P(a_n),$$ is perfectly correct. If you want something that looks more formal, you can express $P(U)$ in as the expectation value of an appropriate operator, the projector $$ \Pi_U=\sum_{n\in J}\sum_i|\varphi_n^i\rangle\langle\varphi_n^i|, $$ but honestly, your expression is perfectly fine.

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The mathematical formalism starts from a C*-algebra $A$, that of observables, and a set of physical states. For any observable $a\in A$, i.e. a self-adjoint element ($a^*=a$) and a state $\omega\in A^*$, the expectation value of $a$ on the state $\omega$ is simply $$\omega(a)$$ If you consider the C*-algebra generated by $a$ you can then apply the Riesz-Markov theorem to obtain a Radon measure $\mu_\omega$ such that $$\omega(a) = \int_{\sigma(a)}\lambda\text d\mu_\omega(\lambda)$$ where $\sigma(a)$ is the spectrum of $a$. Since states $\omega$ are normalised positive functional, $\mu$ is also a probability measure. This allows for the interpretation of the Borel functional calculus $\chi_U(a)$, where $U$ is any Borel subset of $\sigma(a)$, as the projection that, when evaluated on $\omega$, gives the probability that a measure of $a$ on the state $\omega$ falls in the set $U$, i.e. $$\omega(\chi_U(a))=\int_{\sigma(a)}\chi_U(\lambda)\text d\mu_\omega(\lambda)=\int_U\text d\mu_\omega(\lambda).$$ When the spectrum of $a$ is discrete then the measure $\mu$ is atomic and the integral reduces to a sum, i.e. $$\omega(\chi_U(a)) = \sum_{\lambda\in U}\mu_\omega(\{\lambda\}),$$ where each $\mu_\omega(\{\lambda\})$ represents the probability of finding $\lambda$ as an outcome of $a$ on the state $\omega$.

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